Problem: $\dfrac{ 6c + 8d }{ -10 } = \dfrac{ 10c + 5e }{ -5 }$ Solve for $c$.
Multiply both sides by the left denominator. $\dfrac{ 6c + 8d }{ -{10} } = \dfrac{ 10c + 5e }{ -5 }$ $-{10} \cdot \dfrac{ 6c + 8d }{ -{10} } = -{10} \cdot \dfrac{ 10c + 5e }{ -5 }$ $6c + 8d = -{10} \cdot \dfrac { 10c + 5e }{ -5 }$ Reduce the right side. $6c + 8d = -{10} \cdot \dfrac{ 10c + 5e }{ -{5} }$ $6c + 8d = {2} \cdot \left( 10c + 5e \right)$ Distribute the right side $6c + 8d = {2} \cdot \left( {10c} + {5e} \right)$ $6c + 8d = {20}c + {10}e$ Combine $c$ terms on the left. ${6c} + 8d = {20c} + 10e$ $-{14c} + 8d = 10e$ Move the $d$ term to the right. $-14c + {8d} = 10e$ $-14c = 10e - {8d}$ Isolate $c$ by dividing both sides by its coefficient. $-{14}c = 10e - 8d$ $c = \dfrac{ 10e - 8d }{ -{14} }$ All of these terms are divisible by $2$ Divide by the common factor and swap signs so the denominator isn't negative. $c = \dfrac{ -{5}e + {4}d }{ {7} }$